Problem: Let $h(x)=-x^3+3 x^2-4$. For what value of $x$ does $h$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $-4$ (Choice C) C $0$ (Choice D) D $-1$
Solution: We can find the relative extrema (i.e. minima and maxima) of $h$ by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $h$ is $h'(x)=-3x(x-2)$. $h'(x)=0$ for $x=0,2$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=2$. $h$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $(-\infty,0)$ $(0,2)$ $(2,\infty)$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $(-\infty,0)$ $x=-1$ $h'(-1)=-9<0$ $h$ is decreasing $\searrow$ $(0,2)$ $x=1$ $h'\left(1\right)=3>0$ $h$ is increasing $\nearrow$ $(2,\infty)$ $x=3$ $h'(3)=-9<0$ $h$ is decreasing $\searrow$ Now let's look at the critical points: $x$ Before After Verdict $0$ $\searrow$ $\nearrow$ Minimum $2$ $\nearrow$ $\searrow$ Maximum Now we can see that $h$ has a relative maximum at $x=2$.